Question

你好我创建了一个脚本来杀死按年龄排序的进程，但是每次PID都在改变时......我该如何解决这个问题这是我的剧本

 #!/bin/bash
    #Argument = -c check -k kill -l list
usage()
{
cat << EOF
usage: $0 options

This script kills all the processes running and leaves the last one sorted by age running.

OPTIONS:
   -c      checks how many proccess are runnig it needs string argument
   -k      Kill all the processes and leaves just the last sorted by age running
   -l      Show the list of procesess to be killed.
EOF
}

CHECK=
KILL=
LIST=

while getopts "hc:k:l:" OPTION
do
     case $OPTION in
         h)
             usage
             exit 1
             ;;
         c)
             CHECK=$OPTARG
             ps -ef | grep -i $CHECK | wc -l
             ;;

         k)
             KILL=$OPTARG
             T2=$(ps -ef | grep -i "$KILL" | awk '{print $3,$5}' | sort -r +1 | sed 1d |awk '{print $1}')
                for f in $T2; do
                        echo "killing $f"
                        kill $f
                done
             ;;
         l)
             LIST=$OPTARG
             T2=$(ps -ef | grep -i "$LIST" | awk '{print $3,$5}' | sort -r +1 | sed 1d |awk '{print $1}')
                for f in $T2; do
                        echo "PID $f"
                done
               ;;
         ?)
             usage
             exit
             ;;
     esac
done

if [[ -z KILL ]] || [[ -z LIST ]] || [[ -z CHECK ]]
then
     usage
     exit 1
fi

并且我也不明白为什么当我在没有参数的情况下调用脚本时，帮助没有显示

Answer 1

如果另一个程序在被杀死时重新启动它，PID将会改变。这实际上与守护进程很常见。

usage永远不会被调用，因为您正在检查字符串 KILL等是否为空，而不是变量。只需在他们面前添加一个美元符号。

PID每次我试图杀死他们时都会改变

1 个答案: